Direct product of two normal subgroups

Solution 1:

Hints: You will need $AB=G$ to prove surjectivity and $A\cap B=\{e\}$ to prove injectivity.

Since you still didn't fix the issue with $f(b,a)$ and $f(ac)$ being meaningless expressions, let me elaborate on this:

You want a homomorphism $f:A\times B\to G$, so you can put a pair $(a,b)\in A\times B$ in $f$ an get some element $f(a,b)\in G$.

• Writing $f(b,a)$ for $b\in B, a\in A$ implies you could put a pair $(b, a)\in(B\times A)$ in $f$, which is not the case, since $f:A\times B \to G$, $\color{red}{\text{not }f:B\times A\to G}$.
• Writing $f(ac)$ for $a,c\in A$, thus $ac\in A$, implies you could just put an element of $A$ in $f$, which is not the case, since $f:A\times B\to G$, $\color{red}{\text{not }f:A\to G}$.

What you want is for $(a,b), (c,d) \in A\times B$ to have $$ f((a,b)\cdot(c,d)) = f(a,b)\cdot f(c,d),$$ where by definition of the operation in $A\times B$ we have $(a,b)\cdot(c,d)=(ac,bd)\in A\times B$. Now you suggest to define $f$ as \begin{align} f : A\times B &\longrightarrow G, \\ (a,b) &\longmapsto ab. \end{align} Using this definition, we can evaluate both sides of the equation we want to prove for $f$ to be a homomorphism: \begin{align} f((a,b)\cdot(c,d)) &= f(ac, bd) = acbd, \\ f(a,b)\cdot f(c,d) &= ab\cdot cd = abcd. \end{align} Now, since $c\in A$, $b\in B$ where $A,B$ are normal in $G$ and $A\cap B=\{e\}$ you conclude from your previous problem that $$ acbd = a(cb)d = a(bc)d = abcd. $$ Put everything together and you obtain \begin{align} f((a,b)\cdot(c,d)) &= f(ac, bd) = acbd = a(cb)d \\&= a(bc)d = abcd = ab\cdot cd = f(a,b)\cdot f(c,d). \end{align}